Question

A 2.750×10−2M solution of NaCl in water is at 20.0∘C. The sample was created by dissolving a sample of NaCl in water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 999.3 mL . The density of water at 20.0∘C is 0.9982 g/mL.
Calculate the mole fraction of salt in this solution.

Calculate the concentration of the salt solution in percent by mass.

Calculate the concentration of the salt solution in parts per million.

Answer 1

Answer:

Mole fraction of salt is 0.00049.

The concentration of the salt solution in percent by mass is 0.16%.

The concentration of the salt solution in parts per million is 1,610.18 .

Explanation:

Molarity of the NaCl solution = $2.750\times 10^{-2} M$

Moles of NaCl = $n_2$

Volume of the solution = 1.000 L

Molarity=$\frac{\text{Moles of compound}}{\text{Volume of solution(L)}}$

$2.750\times 10^{-2} M=\frac{n_1}{1 L}$

$n_1=2.750\times 10^{-2} mol$

Mass of $2.750\times 10^{-2} mol$ of NaCl :

$2.750\times 10^{-2} mol\times 58.5 g/mol=1.60875 g$

Mass of water = m

Density of water = 0.9982 g/mL

Volume of water = 999.3 mL

$Mass=Density\times Volume$

$m=0.9982 g/mL\times 999.3 mL=997.50 g$

Moles of water =$n_1=\frac{997.5 g}{18 g/mol}=55.416 mol$

Mole fraction of salt = $\chi_1$

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{2.750\times 10^{-2} mol}{2.750\times 10^{-2} mol+55.416 mol}=0.00049$

Percentage by mass:

$\frac{\text{Mass of Solute}}{\text{Mass of Solution}}\times 100$

$\frac{1.60875 g}{1.60875 g+997.50 g}\times =0.16\%$

The concentration of the salt solution in percent by mass is 0.16%.

The concentration of the salt solution in parts per million.

$=\frac{\text{Mass of solute}}{\text{Mass of solution(mL)}}\times 10^6$

$\frac{1.60875 g}{1.60875 g+997.50 g}\times 10^6=1,610.18 ppm$